DEBUG/SntpClient(75): request time faile

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DEBUG/SntpClient(75): request time faile

Postby krishnaveni » Tue Jun 26, 2012 1:55 pm

Hi.,

i have tried to retrieve data from mysql database in android application.
Here as am referred these site:

http://codeoncloud.blogspot.in/2012/03/ ... lient.html

But am not getting output.i have got blank screen only not displayed retrieve data.
Here i got one sntp error on my console window.The error is:

DEBUG/SntpClient(71): request time failed: java.net.SocketException: Address family not supported by protocol.

The following codes are my android application:

The RetailerActivity.java file is

Code: Select all
package com.retailer.client;

import android.app.Activity;
import android.os.Bundle;
import org.ksoap2.SoapEnvelope;
import org.ksoap2.serialization.SoapObject;
import org.ksoap2.serialization.SoapPrimitive;
import org.ksoap2.serialization.SoapSerializationEnvelope;
import org.ksoap2.transport.HttpTransportSE;
import android.widget.TextView;

public class RetailerActivity extends Activity {
    private static final String SOAP_ACTION = "http://ws.retailer.com/customerData";
    private static final String METHOD_NAME = "customerData";
    private static final String NAMESPACE = "http://ws.retailer.com";
    private static final String URL = "http://192.168.1.249:8085/Retailer/services/RetailerWS?wsdl";
    /** Called when the activity is first created. */
    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.main);
        SoapObject request = new SoapObject(NAMESPACE, METHOD_NAME);
       
        SoapSerializationEnvelope envelope = new SoapSerializationEnvelope(SoapEnvelope.VER11);
         
        envelope.setOutputSoapObject(request);
 
        HttpTransportSE ht = new HttpTransportSE(URL);
        try {
            ht.call(SOAP_ACTION, envelope);
            SoapPrimitive response = (SoapPrimitive)envelope.getResponse();
            SoapPrimitive s = response;
            String str = s.toString();
            String resultArr[] = str.split("&");//Result string will split & store in an array
           
            TextView tv = new TextView(this);
       
            for(int i = 0; i<resultArr.length;i++){
            tv.append(resultArr[i]+"\n\n");
           }
            setContentView(tv);
   
        } catch (Exception e) {
            e.printStackTrace();
        }
    }
}


The AndroidManifest.xml file is:

Code: Select all
<?xml version="1.0" encoding="utf-8"?>
<manifest xmlns:android="http://schemas.android.com/apk/res/android"
    package="com.retailer.client"
    android:versionCode="1"
    android:versionName="1.0" >

    <uses-sdk android:minSdkVersion="8" />
    <uses-permission android:name="android.permission.INTERNET"/>

    <application
        android:icon="@drawable/ic_launcher"
        android:label="@string/app_name" >
        <activity
            android:name=".RetailerActivity"
            android:label="@string/app_name" >
            <intent-filter>
                <action android:name="android.intent.action.MAIN" />

                <category android:name="android.intent.category.LAUNCHER" />
            </intent-filter>
        </activity>
       
    </application>


</manifest>


The main.xml file is:

Code: Select all
<?xml version="1.0" encoding="utf-8"?>
<LinearLayout xmlns:android="http://schemas.android.com/apk/res/android"
    android:layout_width="fill_parent"
    android:layout_height="fill_parent"
    android:orientation="vertical" >

    <TextView
        android:layout_width="fill_parent"
        android:layout_height="wrap_content"
        android:text="@string/hello" />

</LinearLayout>


How can i resolve dis error.Please guide me.
krishnaveni
Junior Developer
Junior Developer
 
Posts: 10
Joined: Wed Dec 21, 2011 6:23 am

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